Calculating the Golden Ratio

Here I’ll exhibit a method from approximating another well known irrational. The Golden Ratio, often denoted , has fascinated people for many years, some going as far as to attribute mystical connotations to it.

crops up a lot in nature and it’s always a pleasure to see it. It’s most easily understood as the positive root of the quadratic , making it equal to . Note that this means it has the property . This time we’ll construct a sequence converging to using the function .

Let , and for all let . Then the first few numbers in this sequence are

One might begin to suspect at this point that , where is the Fibonacci sequence (with ). We can prove this is the case by induction, we’ve already shown the base case by listing the first few values above. Then if we assume that for some , then

.

This sequence is known to converge to , but I’ve never explicitly convinced myself of this before so I’ll do that now. First we’ll define a new sequence by . In doing so we’ve ditched the first element of the sequence and so now for all , . This is because for all we have that implies , which you can check out for yourself. Now we note that for all we have

,

since achieves a maximum of on the square .

(This shows that is a contraction mapping on the interval . At this point we could appeal to the Banach fixed-point theorem and we’d be done. Instead, we directly prove convergence of our series using some inspiration from that theorem).

The next thing we can so is derive the following inequality (from here on in we’re assuming that all numbers we consider are between 1.5 and 2):

Clearly

.

Now if we assume that for some , then

.

Thus by induction for all .

Now we show this sequence is Cauchy convergent by noting that for all

which clearly approaches as and go to . So (and thus ) is Cauchy convergent, and thus convergent since is complete. To see that it converges to , we’ll use the continuity of :

Since the equation has a unique solution for , it must be that .

Thus by iterated application of the function we may approximate . Continued iteration yields ever more accurate approximations.

Let me know if I’ve made any mistakes, I haven’t given myself much proof-reading time before publishing (=